How will you prove the formula sin(AB)=sinAcosBcosAsinB using formula of scalar product of two vectors?

1 Answer
Jun 27, 2016

See below

Explanation:

Consider two vectors representing parallelogram sides

a={0,cos(A),sin(A)}
b={0,cos(B),sin(B)}

We know that the parallelogram area is given by

S=abcos(BA)=a,b

then

cos(BA)=a,bab

substituting we obtain

cos(BA)=cos(A)cos(B)+sin(A)sin(B)

We can also obtain the parallelogram area using the cross product.

S=absin(BA)=ˆi,a×b=cos(A)sin(B)cos(B)sin(A)

so

sin(BA)=cos(A)sin(B)cos(B)sin(A)

Here a=b=1

Note.
a,b={0,cos(A),sin(A)},{0,cos(B),sin(B)}=cos(A)cos(B)+sin(A)sin(B)
a×b={0,cos(A),sin(A)}×{0,cos(B),sin(B)}={cos(A)sin(B)cos(B)sin(A),0,0}